How do you simplify #(4 + 4i ) div(5 + 4i )#?

Redirected from "Question #f8633"
1 Answer
Mar 9, 2016

#36/41 +4/41 i #

Explanation:

To simplify the fraction , require to make the denominator real.
This is achieved by multiplying the complex number on the denominator by it's #color(blue)" complex conjugate "#

If (a + bi ) is a complex number then it's conjugate is (a - bi )

Note that the 'real part' remains unchanged , while the 'imaginary' part becomes negative.

and (a+bi)(a-bi) #= a^2-bi+bi-bi^2 = a^2+b^2" a real number "#

using #[ i^2 = (sqrt(-1))^2 = -1 ]#

now multiply numerator and denominator by (5 - 4i )

#rArr( (4+4i)(5-4i))/((5+4i)(5-4i)) = (20-16i+20i-16i^2)/(25-16i^2)#

which simplifies to

#(36+4i)/41 = 36/41 + 4/41 i #